
%!TEX program = xelatex
%!TEX TS-program = xelatex
%!TEX encoding = UTF-8 Unicode

\documentclass[10pt]{article} 

\input{wang_preamble.tex}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{titling}
\setlength{\droptitle}{-2cm}   % This is your set screw

%%文档的题目、作者与日期
\author{王立庆（2022级数学与应用数学1班）}
%\author{ALEX }
\title{高等代数(一)复习 14 - 26}
%\date{\vspace{-3ex}}
\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
\date{2022 年 12 月 29 日}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}

\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %1
下述哪个不是实数域上的向量空间？
\begin{enumerate}
\item 复数全体，复数的加法，以及实数和复数的乘法。
\item 长度为 $n$ 的实数数组全体，按分量相加，按分量相乘。
\item 实系数齐次线性方程组的解集，通常的加法与数乘。
\item 实系数非齐次线性方程组的解集，通常的加法与数乘。
\end{enumerate}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\color{red}解答：(d). 非齐次线性方程组的解集不在加法或数乘下封闭。

}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %2
在向量空间中，数量零乘以任意向量的结果都是零向量。证明可以写成以下几步。其中不正确的是哪个？

\begin{enumerate}
\item 记 $\beta = 0\cdot \alpha$. 
\item 由分配律可得 $\beta+\beta =0\cdot \alpha + 0\cdot \alpha = (0+0) \cdot \alpha = 0\cdot \alpha = \beta$. 
\item 由零向量存在公理，存在向量 $\gamma$ 使得 $\gamma+\beta=\theta$, 其中 $\theta$ 为零向量。
\item 使用加法结合律，由 $\gamma+(\beta+\beta) = \gamma+\beta$ 可得 $\beta =\theta$. 
\end{enumerate}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\color{red}解答：(c). 由负向量存在公理。

}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %3
设 $A$ 是一个 $m\times n$ 阶的实数矩阵，
设 $W$ 是齐次线性方程组 $AX=0$ 的解集。
设 $U$ 是矩阵 $A$ 的行向量组线性张成的子空间。
下述说法中，不正确的是哪个？

\begin{enumerate}
\item  $W$ 是 $\mathbb{R}^n$ 的子空间，维数为 $n-R(A)$. 
\item  $U$ 是 $\mathbb{R}^n$ 的子空间，维数为 $R(A)$. 
\item  $W$ 与 $U$ 的交集只含有零向量。
\item  $U$ 也是矩阵 $A$ 的列向量组线性张成的子空间。
\end{enumerate}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\color{red}解答：(d). 矩阵 $A$ 的行空间是 $\mathbb{R}^n$ 的子空间，列空间 是 $\mathbb{R}^m$ 的子空间。

}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %4
设 $\alpha_1=(1,2,3)$, $\alpha_2=(4,5,6)$, 设 $W=L(\alpha_1,\alpha_2)$ 是由这两个向量线性张成的子空间。
下述说法中，不正确的是哪个？

\begin{enumerate}
\item 向量 $(7,8,9)\in W$. 
\item 向量 $(4,3,2)\in W$. 
\item 向量 $(1,1,1)\in W$. 
\item 向量 $(3,6,5)\in W$. 
\end{enumerate}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\color{red}解答：(d). 由 $\alpha_1,\alpha_2$ 不成比例，可知 $W$ 是一个平面。加入第三个向量，计算三阶行列式。若行列式不为零，则这三个向量仍线性无关，此时第三个向量不在前两个向量张成的平面中。

}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %5
哪个不是下述向量组的极大线性无关组， 
$$\Phi = \{ \alpha_1=(1,1,1,1),  \alpha_2=(1,2,1,2),  \alpha_3=(1,2,2,1),  \alpha_4=(1,2,0,3)\}. $$

\begin{enumerate}
\item $\{ \alpha_1, \alpha_2, \alpha_3 \}$. 
\item $\{ \alpha_1, \alpha_2, \alpha_4 \}$. 
\item $\{ \alpha_1, \alpha_3, \alpha_4 \}$. 
\item $\{ \alpha_2, \alpha_3, \alpha_4 \}$. 
\end{enumerate}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\color{red}解答：(d). 按列向量的方式排列成一个矩阵，使用初等变换化为行最简形，可得 $\{ \alpha_1, \alpha_2, \alpha_3 \}$ 线性无关，而且 $\alpha_4 = 2\alpha_2-\alpha_3$. 

}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %6
关于向量空间的基，下述说法中，不正确的是哪个？
\begin{enumerate}
\item 向量空间的基是这个向量空间中的一组有序向量。 
\item 向量空间的基是一个线性无关的向量组。
\item 向量空间的每个向量都可以由它的一个基通过线性组合得到。
\item 每个向量空间都存在唯一的一个基。
\end{enumerate}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\color{red}解答：(d). 除了零空间，每个向量空间都存在无穷多个基。

}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %7
记 $V=\mathbb{R}[x]_3$ 为次数不超过3的实系数多项式全体组成的集合。
则 $V$ 在多项式的加法与数乘运算下成为一个实向量空间。
求向量 $\xi=(x+1)^2(x+2)$ 关于基 $(1,x,x^2,x^3)$ 的坐标。

\begin{enumerate}
\item $(1,4,5,2)$. 
\item $(2,5,4,1)$. 
\item $(1,2,1,2)$. 
\item $(2,1,2,1)$. 
\end{enumerate}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\color{red}解答：(b). $\xi=(x+1)^2(x+2) = 2+5x+4x^2+x^3$. 

}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %8
设 $f:V\to W$ 是向量空间的同构。下述说法中，不正确的是哪个？
\begin{enumerate}
\item  $f$ 将 $V$ 中的单位向量对应到 $W$ 中的单位向量。
\item  对任意 $\alpha,\beta\in V$, 有 $f(\alpha+\beta) = f(\alpha)+f(\beta)$. 
\item  向量组 $\Phi=\{\alpha_1,\cdots,\alpha_n\}$ 线性相关当且仅当向量组 $f(\Phi)$ 线性相关。
\item  逆映射 $f^{-1}:W\to V$ 也是同构。
\end{enumerate}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\color{red}解答：(a). 向量空间里没有单位向量的概念。

}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %9
行初等变换不改变矩阵的列向量组的线性关系。
下述证明中，不正确的是那一句？

\begin{enumerate}
\item  设 $A$ 是一个 $m\times n$ 矩阵。写成按列向量分块的形式为 $A=(\beta_1, \cdots, \beta_n)$. 
\item  记 $K=(k_1,\cdots,k_n)$, 线性关系 $k_1\beta_1+\cdots+k_n\beta_n=0$ 可以写成 $AK=0$. 
\item  对矩阵 $A$ 进行行初等变换得到的矩阵可以写成 $PA$ 的形式，其中 $P$ 是一个可逆矩阵。
\item  若 $P$ 是可逆矩阵，则 $AK=0$ 等价于 $PAK=0$. 因此 $PA$ 的列向量组也有这个线性关系。
\end{enumerate}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\color{red}解答：(b). 线性关系的系数 $(k_1,\cdots,k_n)$ 需要写成列向量的形式，即 $K=(k_1,\cdots,k_n)^t$, 才能将 $k_1\beta_1+\cdots+k_n\beta_n=0$ 写成 $AK=0$. 

}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %10
设 $A$ 是一个 $m\times n$ 矩阵。设 $R(A)=n-2$. 设 $\alpha_1, \alpha_2, \alpha_3$ 是非齐次线性方程组 $AX=b$ 的三个解向量，
且 $\{\alpha_1, \alpha_2, \alpha_3\}$ 线性无关。下述说法中，不正确的是哪个？

\begin{enumerate}
\item $\{\alpha_1-\alpha_2, \alpha_1-\alpha_3\}$ 是 $AX=0$ 的一个基础解系。
\item $\{\alpha_1-\alpha_2, \alpha_2-\alpha_3\}$ 是 $AX=0$ 的一个基础解系。
\item $AX=b$ 的通解可以写成 $k\alpha_1+m\alpha_2 - (k+m-1)\alpha_3$, 其中 $k,m$ 为任意实数。
\item $AX=b$ 的通解可以写成 $k\alpha_1+m\alpha_2 +n\alpha_3$, 其中 $k,m,n$ 为任意实数。
\end{enumerate}

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\color{red}解答：(d). 齐次线性方程组 $AX=0$ 的基础解系含有 $n-R(A)=2$ 个向量。

}

\vspace{0.2cm}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\end{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
